What is the solution for y^'=sqrt(y+sin(x))-cos(x) ?

The solution for y^'=sqrt(y+sin(x))-cos(x) is y=(x^2+c_{1}^2)/4+(c_{1}x)/2-sin(x)

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y^'=sqrt(y+sin(x))-cos(x)

y^{^{'}} = y + sin (x) - cos (x)

y = \frac{x ^{2} + c _{1}^{2}}{4} + \frac{c _{1} x}{2} - sin (x)

Solution steps

y^{'} (x) = y + sin (x) - cos (x)

Solve using substitution:

y = \frac{x ^{2} + c _{1}^{2}}{4} + \frac{c _{1} x}{2} - sin (x)

y = \frac{x ^{2} + c _{1}^{2}}{4} + \frac{c _{1} x}{2} - sin (x)

f (x) = \frac{7 x - 5}{x ^{2} + 4 x}

f (x) = x^{4} + 4 x^{3} - 48 x^{2} + 6

\int_{0}^{9} 2 x d x

\int_{0}^{\frac{π}{3}} sec^{2} (x) d x

- 4 x^{2} + 12 x, 2 x^{2} - 5

What is the solution for y^'=sqrt(y+sin(x))-cos(x) ?
The solution for y^'=sqrt(y+sin(x))-cos(x) is y=(x^2+c_{1}^2)/4+(c_{1}x)/2-sin(x)